2 2. (1pt each) Answer the following either True or False and justify your answer. (a) If {v 1, v 2, v 3 } is a linearly independent set, then so is the set {kv 1, kv 2, kv 3 } for every nonzero scalar k Solution: TRUE. Since {v 1, v 2, v 3 } is linearly independent, the only way to write the zero vector as a linear combination of v 1, v 2, and v 3 is 0v 1 + 0v 2 + 0v 3 = 0. Consider writing the zero vector as a linear combination of {kv 1, kv 2, kv 3 }. That is, what c 1, c 2, and c 3 satisfy c 1 kv 1 + c 2 kv 2 + c 3 kv 3 = 0. Dividing both sides of this equation by k results in c 1 v 1 + c 2 v 2 + c 3 v 3 = 0. Since the set {v 1, v 2, v 3 } is linearly independent, we know that c 1 = c 2 = c 3 = 0. Hence the set {kv 1, kv 2, kv 3 } is linearly independent. (b) If A x = b does not have any solutions, then b is not in the column space of A. Solution: TRUE. By theorem, a solution to A x = b exists if and only if the vector b can be written as a linear combination of the columns of A. An equivalent statement of the theorem is that a solution to A x = b exists if and only if the vector b lies in the column space of A. (c) There are three linearly independent vectors in R 2. Solution: FALSE. One solution is to note that any basis of R 2 has only two vectors since the dimension of R 2 is 2. Therefore, any set of three vectors can not be a basis and the set has to be linearly dependent. Another solution would be to remember that the homogeneous system ([ v 1 v 2 v 3 0]) formed from three vectors in R 2 that has only two rows. A homogeneous system of less equations

3 than unknowns has to have parametric solutions. Therefore the set of vectors is linearly dependent. Another solution would to draw a picture similar to Example 1(c) in Lesson 4.5. (d) Every nonzero subspace of R n has a unique basis. Solution: FALSE. Bases are not unique The number of vectors in a basis of a subspace is unique, but the vectors chosen for the basis are not unique. If the subspace is n-dimensional, any selection of n linearly independent vectors from the subspace will span the subspace. (e) Let A be a 3 3 matrix. Then the adjoint of A is the (i, j) cofactor. A 11 A 12 A 13 A 21 A 22 A 23, where A ij is A 31 A 32 A 33 A 11 A 21 A 31 Solution: FALSE. The definition of the adjoint of A is adj A = A 12 A 22 A 32. A 13 A 23 A 33

4 3. (a) (5pts) Define a subspace W of a vector space V. Solution: A subset W of V is a subspace of V if it is closed under V s operations of vector addition and scalar multiplication. (b) (5pts) Show that the set S of all vectors of the form (a, a + b, b) for all real numbers a and b is a subspace of R 3. Solution: One way to show that this is a subspace is to show that the set of vectors (a, a+b, b) is the span of a set of vectors. (a, a + b, b) = (a, a, 0) + (0, b, b) for all a and b = a(1, 1, 0) + b(0, 1, 1) for all a and b = span {(1, 1, 0), (0, 1, 1)} by def n of a span Since every span is a subspace, the set S is a subspace. Another way to show that this is a subspace is to show that the set is closed under addition and scalar multiplication. Let u = (a 1, a 1 + b 1, b 1 ) and v = (a 2, a 2 + b 2, b 2 ) be two different vectors in R 2. closed under addition u + v = (a 1 + a 2, a 1 + b 1 + a 2 + b 2, b 1 + b 2 ) = (a 1 + a 2, a 1 + a 2 + b 1 + b 2, b 1 + b 2 ) = ((a 1 + a 2 ), (a 1 + a 2 ) + (b 1 + b 2 ), (b 1 + b 2 )) which is still a vector of the form of the set. Therefore, the set S is closed under addition. closed under scalar multiplication

6 4. (a) (3pts) Define a basis for a vector space V. Solution: A set of vectors is a basis for a vector space V if the set of vectors is linearly independent and the span of the set of vectors is all of V. (b) (5pts) Use a dependency table and find a basis for the column space of A = Solution: First, we create the initial dependency table using matrix A: e 1 e 2 e 3 a 1 a 2 a 3 a 4 a Moving the matrix into reduced row echelon form, we get: a 1 a 3 a 5 a 1 a 2 a 3 a 4 a We see that the set { a 1, a 3, a 5 } form a basis for column space. (c) (2pts) What is the dimension of the column space of A? Explain your answer. Solution: Since there are three vectors in the basis, we know that the dimension of the columns space of A is three.

7 5. (5pts) Let u and v be vectors in a vector space. Under what conditions will it be true that span(u) = span(v). Solution: We know that the span of a single vector forms a line. If span(u) = span(v), then the spans define the same line. In other words, the line x = tu for all t has to be the same line as x = sv for all s. Therefore, u and v have to be scalar multiples of each other. Another decent answer would be to discuss Example 1(e) is Lesson Let A be a 4 4 matrix with det A = 2. (a) (2pts) Does A 1 exist? Explain your answer. Solution: Since the determinant of A is nonzero, we know that A 1 exists. (b) (3pts) Can the linear system A x = b have more than one solution? Explain your answer. Solution: Since A is invertible, we know that the reduced row echelon form of A is the identity matrix I. Therefore [A b] has the reduced row echelon form of [I p] where p is the unique solution point to the linear system.

8 7. (a) (5pts) Define what it means for a vector to be the additive identity element of the vector space. Solution: The additive identity element is the vector of a vector space that satisfies property 4 of the definition of a vector space. That is, it is the vector 0 such that for any other vector v in the vector space, v + 0 = 0 + v = v. (b) (5pts) Show that it is not possible for a vector space to have two different zero vectors. That is, is it possible to have two different vectors 0 1 and 0 2 such that these vectors both satisfy the fourth property a vector space? Explain your reasoning. Solution: This is Homework 6, Problem 4. Assume that there are two different zero vectors 0 1 and 0 2. By the definition of a zero vector, for any v in the vector space v + 0 = v. For 0 1, this amounts to v = v. Let v be the other zero. Then = 0 2. Similarly, for 0 2, this amounts to v = v. Let v be the other zero. Then = 0 1. Since addition is commutative for the vector space, = Therefore, 0 1 = 0 2 ; showing that every vector space has exactly one zero element.

11 2. (1pt each) Answer the following either True or False and justify your answer. (a) If S is a finite set of vectors in a vector space V, then span S must be closed under vector addition and scalar multiplication. Solution: TRUE. The span of a set of vectors is a subspace. By definition, a subspace is closed under vector addition and scalar multiplication. (b) If span(s 1 ) = span(s 2 ) then S 1 = S 2. Solution: FALSE. The spanning sets describe the same subspace but that does not mean that the exact same set of vectors are are in each spanning set. For example, if S 1 and S 2 were bases (which they don t have to be in this question), we know that even bases are not unique. (c) If {v 1, v 2 } is a linearly dependent set of nonzero vectors, then each vector is a scalar multiple of the other. Solution: TRUE. By the definition of a linear dependent set, there exist constants c 1 and c 2 such that c 1 v 1 + c 2 v 2 = 0. This shows that either vector can be written as a scalar multiple of the other. For example, v 1 = c 2 c 1 v 2. (d) No set of two vectors can span R 3. Solution: TRUE. Since R 3 is 3-dimensional, a set of vectors that span all of R 3 has to have at least 3 vectors. (e) If A, B, and C are square matrices, then det ABC = det A det B det C. Solution: TRUE. Recall that we have the theorem det (AB) = det A det B To show this

13 3. (a) (3pts) Define what it means for a vector v to be a linear combination of a set of vectors. Solution: A vector v is a linear combination of a set of vectors if v can be written as the sum of scalar multiples of the set. In other words, if v is a linear combination of the set {v 1, v 2,..., v k } then v = c 1 v 1 + c 2 v c k v k for some constants c i. (b) (5pts) In R 3, determine if b = (5, 1, 1) can be written as a linear combination of the vectors (1, 9, 1), ( 1, 3, 1), and (1, 1, 1). If so, write the combination. Solution: We are determining if there are constants c 1, c 2 and c 3 such that c 1 (1, 9, 1) + c 2 ( 1, 3, 1) + c 3 (1, 1, 1) = (5, 1, 1). Forming the augmented matrix, The reduced row echelon form of this matrix is Therefore, the constants are c 1 = 1, c 2 = 3 and c 3 = 1 and b can be written as the combination b = (1, 9, 1) 3( 1, 3, 1) + (1, 1, 1). (c) (2pts) Without doing any computations, explain if b = (5, 1, 1) lies in the column space of Solution: The column space of is the span {(1, 9, 1), ( 1, 3, 1), (1, 1, 1)}. Above we have shown that b is a linear combination of these vectors. Therefore, b lies in the column space.

14 4. (a) (5pts) List 4 of the 10 properties in the definition of a real vector space. Solution: Let V be the vector space, u and v vectors in V and c and d be any real numbers. Then any four of the following will work: 1. V is closed under vector addition 2. addition is commutative 3. addition is associative 4. V has an additive identity element 5. every element has an additive inverse 6. V is closed under scalar multiplication 7. c(u + v) = cu + cv 8. (c + d)u = cu + du 9. c(du) = (cd)u 10. 1u = u (b) (5pts) Under the normal operations of addition and scalar multiplication, show that the set of integers is not a real vector space. Solution: The set of integers {..., 3, 2, 1, 0, 1, 2, 3,...} are not closed under scalar multiplication. Let n be any integer. Then kn does not have to be an integer. For example, if k = π then πn is not an integer.

15 5. (5pts) Under what conditions will two vectors in R 3 span a plane? A line? Clearly explain your answer. Solution: Let v 1 and v 2 be vectors in R 3. If the two vectors are linearly independent then the set {v 1, v 2 } form a basis for the subspace spanned by the set {v 1, v 2 }. Two basis vectors mean that the subspace is 2-dimensional. Hence, if the two vectors are linearly independent the span of them forms a plane. If v 1 and v 2 are linearly dependent, then span {v 1, v 2 } = span {v 1 } = span {v 2 } since v 1 = kv 2 for some constant k. The span of a single vector forms a line. 6. (5pts) Show that it is not possible for a vector u in a vector space to have two different negatives. That is, for the vector u, it is not possible to have two different vectors ( u) 1 and ( u) 2. Clearly justify your answer. Solution: This is Homework 6, Problem 5. Assume u does have two different additive inverses. By the definition of inverses, we have the two equations u + ( u) 1 = 0 and u + ( u) 2 = 0. Therefore u + ( u) 1 = u + ( u) 2. On the left-hand side of both sides of the equation, add one of u s negatives. (It doesn t matter which.) Then ( u) 1 + u + ( u) 1 = ( u) 1 + u + ( u) 2. [( u) 1 + u] + ( u) 1 = [( u) 1 + u] + ( u) 2 by associativity. [0] + ( u) 1 = [0] + ( u) 2 by the properties of negatives. ( u) 1 = ( u) 2 by the properties of zeros. Therefore every element can have only one unique additive inverse.

16 7. Let v 1 = ( 2, 0, 1), v 2 = (3, 2, 5), v 3 = (6, 1, 1), and v 4 = (7, 0, 2). (a) (3pts) What vector space are these vectors elements of? Explain your answer. Solution: Since every vector is an 3-tuple of numbers, the vectors are elements of R 3. (b) (4pts) Show that the set {v 1, v 2, v 3, v 4 } is linearly dependent. Solution: One solution would be to site the theorem that says that since we have more vectors (4) than dimensions of the the vector space (3), the four vectors have to be linearly dependent. This is a fine answer to part (b) but won t help us with (c). I will show dependency using the definition. We want to show that there exists a linear combination of the set, without all zero scalars, that forms the zero vector. Solving for the coefficients equates to the linear system [v 1 v 2 v 3 v 4 0]. Forming the augmented matrix, The reduced row echelon form of this matrix is / / /29 0. Since the system has parametric solutions, the set is linearly dependent. (c) (3pts) Find a dependency equation. Solution: Solving for the coefficients in the above reduced row echelon form, we get c 1 = t, c 2 = 3 29 t, c 3 = 6 29 t, and c 4 = t. Therefore, 0 = tv tv tv 3 + tv 4 for all t. To find a dependency equation, the easiest way is to let t = 1. This yields the dependency equation, 0 = v v v 3 + v 4.

17 8. Let A be a 4 4 matrix with det A = 2. (a) (3pts) If A is transformed into the reduced row echelon form matrix B, what is B? Explain your answer. Solution: Since the determinant is nonzero, we know that the matrix A is invertible. Since A is invertible, we know that the reduced row echelon form of the matrix is the identity matrix. (b) (2pts) Describe the set of all solutions to the homogeneous system A x = 0. Solution: Since the matrix is invertible, we know that the system has a unique solution. The only unique solution to the null space problem is the trivial solution (0, 0, 0, 0).

18 9. (5pts) Sink Inc. manufactures sinks from a steel alloy that is 76% iron, 16% nickel, and 8% chromium and counter tops from an alloy that is 70% iron, 20% nickel, and 10% chromium. They have located two supplies of scrap metal and can save substantially on the production costs if their products can be blended from these scraps. One scrap metal mixture contains 80% iron, 15% nickel, and 5% chromium, and the other scrap metal mixture contains 60% iron, 20% nickel, and 20% chromium. Determine if the products can be manufactured from a blend of these scrap mixtures; if so, find the blend. Solution: Interpreting the metal composition from the sources and the desired alloys as vectors, we can write the above information in the following form: S 1 = (80, 15, 5), S 2 = (60, 20, 20), A 1 = (76, 16, 8), and A 2 = (70, 20, 10). To determine if the alloys can be made from the sources, we look at the matrix [ S 1S2 A 1A2 ] Putting this matrix into reduced row echelon form, we get Note that only the Alloy 1 vector is consistent. Therefore, from these two sources, we can only make Alloy 1. The conclusion: To make Alloy 1 (and subsequently the sinks), use a combination of metal that consists of 80% of the metal being from Source 1 and 20% from Source 2.

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